# Multiplying a fraction

## Homework Statement

I would just like to know if I got this right. ((x2 + 2x + 1)/(18cw3) times ((12c3w)/(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3)(x2 - 18cw3

## Homework Equations

((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))

## The Attempt at a Solution

(12c3wx2 + 2x12c3w + 12c3w)/((18cw3)(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3x2 - 18cw3

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Samy_A
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Removed, as apparently there is an error in the formulation of the question.

I accidentally used a subscript all the way through instead of a power, argh!!

Mark44
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##(12c_3wx_2 + 2x12c_3w + 12c_3w)/(18cw_3x_2 - 18cw_3##
On the right side one of the terms is ##2x12c_3w##. Should that have been ##2x^{12}c^3##? Also, you're missing a right parenthesis.

haruspex
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((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))
Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.

On the right side one of the terms is ##2x12c_3w##. Should that have been ##2x^{12}c^3##? Also, you're missing a right parenthesis.
It should have been as haruspex said: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1)

Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.
Yes, that is what I meant.

Yes, that is what I meant.
Do you mean ##\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}##?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.
(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)

haruspex
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(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)
More cancellation to go. Look at the constants, look at the powers of c and w.

More cancellation to go. Look at the constants, look at the powers of c and w.
(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)

(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)
The 3w is canceled

The 3w is canceled
I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)

haruspex
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I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)
You made a mistake with the constants.

You made a mistake with the constants.
Should it be: 3cw(w2) and 3cw(4c2)??

Should it be: 3cw(w2) and 3cw(4c2)??
Oops I meant: 3cw(6w2) and 3cw(4c2)

haruspex
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Oops I meant: 3cw(6w2) and 3cw(4c2)
Yes.

Yes.

Yes.
So the answer is: (x+1)/(6w2) times (4c2)/(x-1)

Mark44
Mentor
(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)
The 3w is canceled
I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)
Should it be: 3cw(w2) and 3cw(4c2)??
Oops I meant: 3cw(6w2) and 3cw(4c2)
Instead of adding new posts each time you discover something you should have said, you can edit the post you want to change.

So the answer is: (x+1)/(6w2) times (4c2)/(x-1)
The constants can be further simplified, and then you should bring it together so you have one fraction.

The constants can be further simplified, and then you should bring it together so you have one fraction.
Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?

Samy_A
Homework Helper
Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?
1)
The constants can be further simplified
2) I think you can leave the nominator as ##2c²(x+1)##, and similarly for the denominator. Certainly would give a better readable result.

Last edited:
science_rules
1)

2) I think you can leave the nominator as ##2c²(x+1)##, and similarly for the denominator. Certainly would give a better readable result.
How did you get 2c2(x+1) when it was 4c2/(x+1)?? Isn't the numerator supposed to be 2c2(2x+2)??
And the denominator is supposed to be: 2w2(3x-3)??

haruspex
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How did you get 2c2(x+1) when it was 4c2/(x-1)?? Isn't the numerator supposed to be 2c2(2x+2)??
And the denominator is supposed to be: 2w2(3x-3)??
You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.

science_rules
You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.
Okay, I think I have it now: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1) = (x+1)2(12c3w)/(18cw2)(x+1)(x-1) Cancel the (x+1) = (12c3w)(x+1)/(18cw2)(x-1) Cancel the c and w = (12c2)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c2)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c2)(x+1)/(3w)(x-1) Answer